$ C = \left[\begin{array}{rrr}4 & -2 & 3 \\ 2 & 5 & 4\end{array}\right]$ $ D = \left[\begin{array}{rr}3 & 2 \\ 3 & -1 \\ 4 & -1\end{array}\right]$ What is $ C D$ ?
Answer: Because $ C$ has dimensions $(2\times3)$ and $ D$ has dimensions $(3\times2)$ , the answer matrix will have dimensions $(2\times2)$ $ C D = \left[\begin{array}{rrr}{4} & {-2} & {3} \\ {2} & {5} & {4}\end{array}\right] \left[\begin{array}{rr}{3} & \color{#DF0030}{2} \\ {3} & \color{#DF0030}{-1} \\ {4} & \color{#DF0030}{-1}\end{array}\right] = \left[\begin{array}{rr}? & ? \\ ? & ?\end{array}\right] $ To find the element at any row $i$ , column $j$ of the answer matrix, multiply the elements in row $i$ of the first matrix, $ C$ , with the corresponding elements in column $j$ of the second matrix, $ D$ , and add the products together. So, to find the element at row 1, column 1 of the answer matrix, multiply the first element in ${\text{row }1}$ of $ C$ with the first element in ${\text{column }1}$ of $ D$ , then multiply the second element in ${\text{row }1}$ of $ C$ with the second element in ${\text{column }1}$ of $ D$ , and so on. Add the products together. $ \left[\begin{array}{rr}{4}\cdot{3}+{-2}\cdot{3}+{3}\cdot{4} & ? \\ ? & ?\end{array}\right] $ Likewise, to find the element at row 2, column 1 of the answer matrix, multiply the elements in ${\text{row }2}$ of $ C$ with the corresponding elements in ${\text{column }1}$ of $ D$ and add the products together. $ \left[\begin{array}{rr}{4}\cdot{3}+{-2}\cdot{3}+{3}\cdot{4} & ? \\ {2}\cdot{3}+{5}\cdot{3}+{4}\cdot{4} & ?\end{array}\right] $ Likewise, to find the element at row 1, column 2 of the answer matrix, multiply the elements in ${\text{row }1}$ of $ C$ with the corresponding elements in $\color{#DF0030}{\text{column }2}$ of $ D$ and add the products together. $ \left[\begin{array}{rr}{4}\cdot{3}+{-2}\cdot{3}+{3}\cdot{4} & {4}\cdot\color{#DF0030}{2}+{-2}\cdot\color{#DF0030}{-1}+{3}\cdot\color{#DF0030}{-1} \\ {2}\cdot{3}+{5}\cdot{3}+{4}\cdot{4} & ?\end{array}\right] $ Fill out the rest: $ \left[\begin{array}{rr}{4}\cdot{3}+{-2}\cdot{3}+{3}\cdot{4} & {4}\cdot\color{#DF0030}{2}+{-2}\cdot\color{#DF0030}{-1}+{3}\cdot\color{#DF0030}{-1} \\ {2}\cdot{3}+{5}\cdot{3}+{4}\cdot{4} & {2}\cdot\color{#DF0030}{2}+{5}\cdot\color{#DF0030}{-1}+{4}\cdot\color{#DF0030}{-1}\end{array}\right] $ After simplifying, we end up with: $ \left[\begin{array}{rr}18 & 7 \\ 37 & -5\end{array}\right] $